What is the tangent of f(x)=-2x^3 ?

The tangent of f(x)=-2x^3 is y=-6a_{0}^2x+4a_{0}^3

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tangent of f(x)=-2x^3

tangent of

f (x) = - 2 x^{3}

y = - 6 a_{0}^{2} x + 4 a_{0}^{3}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - 2 a_{0}^{3})

Find the slope of

f (x) = - 2 x^{3} : \frac{df ( x )}{d x} = - 6 x^{2}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 6 a_{0}^{2}

Find the line with slope m=

- 6 a_{0}^{2}

and passing through

(a_{0}, - 2 a_{0}^{3}) : y = - 6 a_{0}^{2} x + 4 a_{0}^{3}

y = - 6 a_{0}^{2} x + 4 a_{0}^{3}

\frac{s - 1}{2 s ^{2} + s + 6}

\int_{2}^{4} (2 x^{2} - 3 x + 1) d x

x \to - 2 lim (\frac{3}{x + 2})

\int 1 - u^{2} d u

y^{^{''}} - 8 y^{^{'}} + 16 y = 0, y (0) = 4, y^{^{'}} (0) = \frac{82}{5}

What is the tangent of f(x)=-2x^3 ?
The tangent of f(x)=-2x^3 is y=-6a_{0}^2x+4a_{0}^3